Question

Factorise:
(i) $4x^2+9y^2+16z^2+12xy-24yz-16xz$
(ii) $2x^2+y^z+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

Answer 1

We know that, the algebraic identity:

$(a^2+b^2+c^2+2ab+2bc+2ca)=(a+b+c{)}^2$

(i)

$4x^2+9y^2+16z^2+12xy-24yz-16xz$

Here, $yz$ and $xz$ terms are negative.

i.e., common variable $z$ is in negative form.

$=(2x{)}^2+(3y{)}^2+(-4z{)}^2+2(2x\left)\right(3y)+2(3y\left)\right(-4z)+2(-4z\left)\right(2x)$

$=(2x+3y-4z{)}^2$ [$\because (a^2+b^2+c^2+2ab+2bc+2ca)=(a+b+c{)}^2$]

$=(2x+3y-4z)(2x+3y-4z)$

(ii)

$2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8zx$

Here, $xy$ and $zx$ terms are negative.

i.e., common variable $x$ is in negative form.

$=(-\sqrt{2}x)+(y{)}^2+(2\sqrt{2}z)+2(-\sqrt{2}x\left)\right(y)+2(y\left)\right(2\sqrt{2}z)+2(2\sqrt{2}z\left)\right(y)$

$=(-\sqrt{2}x+y+2\sqrt{2}z{)}^2$ [$\because (a^2+b^2+c^2+2ab+2bc+2ca)=(a+b+c{)}^2$]

$=(-\sqrt{2}x+y+2\sqrt{2}z)(-\sqrt{2}x+y+2\sqrt{2}z)$